problem11_54

University Physics with Modern Physics with Mastering Physics (11th Edition)

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11.54: a) Take torques about the pivot. The force that the ground exerts on the ladder is given to be vertical, and θ θ F sin ) m 0 . 4 )( N 250 ( sin ) m 0 . 6 ( V = N. 354 so , sin ) m 50 . 1 )( N 750 ( V = + F θ b) There are no other horizontal forces on the ladder, so the horizontal pivot force is zero. The vertical force that the pivot exerts on the
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Unformatted text preview: ladder must be N, 646 ) N 354 ( ) N 250 ( ) N 750 ( =-+ up, so the ladder exerts a downward force of N 646 on the pivot. c) The results in parts (a) and (b) are independent of ....
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This document was uploaded on 02/05/2008.

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